Question

Consider a short cylinder of radius \(r_{o}\) and height \(H\) in which heat is generated at a constant rate of \(\dot{e}_{\mathrm{gen}}\). Heat is lost from the cylindrical surface at \(r=r_{o}\) by convection to the surrounding medium at temperature \(T_{\infty}\) with a heat transfer coefficient of \(h\). The bottom surface of the cylinder at \(z=0\) is insulated, while the top surface at \(z=H\) is subjected to uniform heat flux \(\dot{q}_{H}\). Assuming constant thermal conductivity and steady two-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary conditions) of this heat conduction problem. Do not solve.

Short answer

Based on the given heat conduction problem in a cylinder, provide a mathematical formulation including the differential equation and boundary conditions. Mathematical formulation for the problem is as follows: Differential equation: $$ \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial T}{\partial r} \right) + \frac{\partial^2 T}{\partial z^2} = -\frac{\dot{e}_{\mathrm{gen}}}{k} $$ Boundary conditions: 1. Bottom surface insulated at \(z=0\): $$ \frac{\partial T}{\partial z}(z=0) = 0 $$ 2. Top surface with uniform heat flux at \(z=H\): $$ -k \frac{\partial T}{\partial z}(z=H) = \dot{q}_{H} $$ 3. Convection heat transfer at the outer surface, \(r=r_{o}\): $$ h\left(T(r_{o},z)-T_{\infty}\right) = -k \frac{\partial T}{\partial r}(r=r_{o}) $$

Step-by-step solution

Write down the general heat conduction equation

For steady two-dimensional heat transfer in a cylindrical coordinate system, the general heat conduction equation with constant thermal conductivity \((k)\) can be expressed as: $$ \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial T}{\partial r} \right) + \frac{\partial^2 T}{\partial z^2} = -\frac{\dot{e}_{\mathrm{gen}}}{k} $$

Apply boundary conditions

We have three boundary conditions to apply: 1. Bottom surface insulated at \(z=0\): Since the bottom surface is insulated, there is no heat transfer through it. So, the temperature gradient with respect to \(z\) is zero at \(z=0\): $$ \frac{\partial T}{\partial z}(z=0) = 0 $$ 2. Top surface with uniform heat flux at \(z=H\): The given heat flux \(\dot{q}_{H}\) is equal to the heat transferred through conduction, so \(\dot{q}_{H} = -k \frac{\partial T}{\partial z}\), then at \(z=H\): $$ -k \frac{\partial T}{\partial z}(z=H) = \dot{q}_{H} $$ 3. Convection heat transfer at the outer surface, \(r=r_{o}\): At the outer surface, the heat transfer is due to convection, so: $$ h\left(T(r_{o},z)-T_{\infty}\right) = -k \frac{\partial T}{\partial r}(r=r_{o}) $$

Provide the mathematical formulation

Now, we can write down the mathematical formulation for the problem including the differential equation and boundary conditions: Differential equation: $$ \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial T}{\partial r} \right) + \frac{\partial^2 T}{\partial z^2} = -\frac{\dot{e}_{\mathrm{gen}}}{k} $$ Boundary conditions: 1. Bottom surface insulated at \(z=0\): $$ \frac{\partial T}{\partial z}(z=0) = 0 $$ 2. Top surface with uniform heat flux at \(z=H\): $$ -k \frac{\partial T}{\partial z}(z=H) = \dot{q}_{H} $$ 3. Convection heat transfer at the outer surface, \(r=r_{o}\): $$ h\left(T(r_{o},z)-T_{\infty}\right) = -k \frac{\partial T}{\partial r}(r=r_{o}) $$ We have now provided the mathematical formulation for the given heat conduction problem.